JEE Advance - Physics (2023 - Paper 1 Online - No. 7)
Two satellites $\mathrm{P}$ and $\mathrm{Q}$ are moving in different circular orbits around the Earth (radius $R$ ). The heights of $\mathrm{P}$ and $\mathrm{Q}$ from the Earth surface are $h_{\mathrm{P}}$ and $h_{\mathrm{Q}}$, respectively, where $h_{\mathrm{P}}=R / 3$. The accelerations of $\mathrm{P}$ and $\mathrm{Q}$ due to Earth's gravity are $g_{\mathrm{P}}$ and $g_{\mathrm{Q}}$, respectively. If $g_{\mathrm{P}} / g_{\mathrm{Q}}=36 / 25$, what is the value of $h_{\mathrm{Q}}$ ?
$\frac{3 R}{5}$
$\frac{R}{6}$
$\frac{6 R}{5}$
$\frac{5 R}{5}$
Explanation
The formula to calculate the acceleration due to gravity at a height ( h ) from the surface of the Earth is expressed as :
$ g = \frac{G M_e}{(R_e + h)^2} $
To find the ratio of gravitational acceleration at two different heights ( h_P ) and ( h_Q ) above the Earth's surface, use the formula for each and form a ratio :
$ \frac{g_P}{g_Q} = \frac{\frac{G M_e}{(R_e + h_P)^2}}{\frac{G M_e}{(R_e + h_Q)^2}} = \frac{(R_e + h_Q)^2}{(R_e + h_P)^2} $
Given $ h_P = \frac{R_e}{3} $, the ratio simplifies to:
$ \frac{g_P}{g_Q} = \frac{(R_e + h_Q)^2}{(R_e + \frac{R_e}{3})^2} = \frac{36}{25} $
Solving for $ h_Q $ in terms of $ R_e $ yields $ h_Q = \frac{3R_e}{5} $.
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